At 25 $\degree$C, the equilibrium constant K_c for the reaction 2A(aq) B(aq) + C(aq) is 65. If 2.50 mol of A is added to enough water to prepare 1.00 L of solution, what will the equilibrium concentration of A be?

Question

Quizplus · Accepted Answer

The balanced equation for the reaction is: 2A(aq) 11eb16b2_ffc5_4519_984d_9db6bf2d5400_TB7799_11 B(aq) + C(aq) The equilibrium constant expression is: K_c = [B]^2[C]/[A]^2 At equilibrium, let x be the concentration of B and C, and 2.50 - 2x be the concentration of A. Then, K_c = x^2/(2.50 - 2x)^2 Substituting the given value of the equilibrium constant, we have: 65 = x^2/(2.50 - 2x)^2 Solving for x, we get: x = 0.14 M Therefore, the equilibrium concentration of A is: 2.50 - 2x = 2.22 M However, the question asks for the equilibrium concentration of A in units of M, so we divide 2.22 by the total volume of the solution, 1.00 L, to get: [ A ] = 2.22/1.00 = 0.22 M Therefore, the best choice is B, with an equilibrium concentration of A of 0.14 M.

At 25 °\degree° C, the Equilibrium Constant Kc for the Reaction 2A(aq)

At 25 $\degree$ C, the Equilibrium Constant K_c for the Reaction 2A(aq)